One of the simplest examples of a non-spacial category is $\mathrm{End}(\mathrm{Vec}^{\oplus 2})$, the category of $2\times2$ matrices with vector space coefficients. Working over $\mathbb C$ this is a multifusion category with four simple objects: $$ \begin{pmatrix} \mathbb C & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & \mathbb C \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ \mathbb C & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & \mathbb C \end{pmatrix}$$ and the tensor product is just matrix multiplication (with tensor product and direct sum of vector spaces, of course). The monoidal unit is not simple: $$ \mathbb{1} = \begin{pmatrix} \mathbb C & 0 \\ 0 & 0 \end{pmatrix} \oplus \begin{pmatrix} 0 & 0 \\ 0 & \mathbb C \end{pmatrix}.$$ In particular, the endomorphisms of the monoidal unit is the commutative ring $\mathbb C \oplus \mathbb C$ (with componentwise addition and multiplication). Consider the endomorphism $$ h = (\alpha,\beta) \in \mathrm{End}(\mathbb 1)$$ and object $$ A = \begin{pmatrix} 0 & \mathbb C \\ 0 & 0 \end{pmatrix}.$$ Then one of your diagrams evaluates to $\alpha \, \mathrm{id}_A$ and the other one evaluates to $\beta \, \mathrm{id}_A$.

Remark: Write $R = \mathrm{End}(\mathbb 1) = \mathbb C \oplus \mathbb C$. Then it is an associative algebra, and the multifusion category above is the monoidal category of $R$-$R$ bimodules. Categories of bimodules are typically not spacial except when the algebra has trivial (i.e. one-dimensional) centre.

One of the simplest examples of a non-spacial category is $\mathrm{End}(\mathrm{Vec}^{\oplus 2})$, the category of $2\times2$ matrices with vector space coefficients. Working over a field $\mathbb k$, this is a multifusion category with four simple objects: $$ \begin{pmatrix} \mathbb k & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & \mathbb k \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ \mathbb k & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & \mathbb k \end{pmatrix}$$ and the tensor product is just matrix multiplication (with tensor product and direct sum of vector spaces, of course). The monoidal unit is not simple: $$ \mathbb{1} = \begin{pmatrix} \mathbb k & 0 \\ 0 & 0 \end{pmatrix} \oplus \begin{pmatrix} 0 & 0 \\ 0 & \mathbb k \end{pmatrix}.$$ In particular, the endomorphisms of the monoidal unit is the commutative ring $\mathbb k \oplus \mathbb k$ (with componentwise addition and multiplication). Consider the endomorphism $$ h = (\alpha,\beta) \in \mathrm{End}(\mathbb 1)$$ and object $$ A = \begin{pmatrix} 0 & \mathbb k \\ 0 & 0 \end{pmatrix}.$$ Then one of your diagrams evaluates to $\alpha \, \mathrm{id}_A$ and the other one evaluates to $\beta \, \mathrm{id}_A$.

Remark: Write $R = \mathrm{End}(\mathbb 1) = \mathbb k \oplus \mathbb k$. Then it is an associative $\mathbb k$-algebra, and the multifusion category above is the monoidal category of $R$-$R$ bimodules. Categories of bimodules are typically not spacial except when the algebra has trivial (i.e. one-dimensional) centre. A fun example to think through is to consider $\mathbb C$ as an $\mathbb R$-algebra. Then the category of $\mathbb C$-$\mathbb C$ bimodules has two simple objects (Exercise: why?), and is not spacial (Exercise: why not?).